Problem 109
Question
Evaluate the definite integrals. $$ \int_{0}^{1} \frac{1}{1+x^{2}} d x $$
Step-by-Step Solution
Verified Answer
The integral evaluates to \( \frac{\pi}{4} \).
1Step 1: Recognize the Integrand
Identify the form of the integrand \( \frac{1}{1+x^2} \). This function is known to correspond to the derivative of the inverse tangent function, i.e., \( \frac{d}{dx} (\tan^{-1}(x)) = \frac{1}{1+x^2} \).
2Step 2: Antiderivative of the Function
Use the recognition from Step 1 to find the antiderivative of the integrand. Since \( \frac{d}{dx}(\tan^{-1}(x)) = \frac{1}{1+x^2} \), the antiderivative is \( \tan^{-1}(x) \).
3Step 3: Apply the Fundamental Theorem of Calculus
By the Fundamental Theorem of Calculus, evaluate the definite integral by calculating \( F(b) - F(a) \), where \( F(x) = \tan^{-1}(x) \), \( a = 0 \), and \( b = 1 \).
4Step 4: Calculate the Antiderivative at the Boundaries
Compute \( \tan^{-1}(1) \). The value is \( \frac{\pi}{4} \), since \( \tan(\frac{\pi}{4}) = 1 \). Now compute \( \tan^{-1}(0) \), which is \( 0 \), as \( \tan(0) = 0 \).
5Step 5: Calculate the Value of the Definite Integral
Subtract the values obtained at the boundaries: \( \frac{\pi}{4} - 0 = \frac{\pi}{4} \).
6Step 6: Conclusion: Result of the Integral
Therefore, the value of the definite integral \( \int_{0}^{1} \frac{1}{1+x^2} \, dx \) is \( \frac{\pi}{4} \).
Key Concepts
Inverse Trigonometric FunctionsFundamental Theorem of CalculusAntiderivatives
Inverse Trigonometric Functions
Inverse trigonometric functions are functions that reverse the action of the standard trigonometric functions like sine, cosine, and tangent. They can be used to find angles when the values of trigonometric ratios are known. For instance:
The inverse tangent function, \( \tan^{-1}(x) \), is particularly important when dealing with certain integrals. As we found from the exercise, the derivative of the inverse tangent function is \( \frac{1}{1+x^2} \). This means, to solve the definite integral \( \int_{0}^{1} \frac{1}{1+x^2} \, dx \), we essentially reverse the differentiation process by identifying that this function is the derivative of \( \tan^{-1}(x) \). Understanding these inverse relationships helps in evaluating integrals that might otherwise seem complex.
- Arcsine, written as \( \sin^{-1}(x) \), finds the angle when the sine is known.
- Arccosine, written as \( \cos^{-1}(x) \), finds the angle when the cosine is known.
- Arctangent, written as \( \tan^{-1}(x) \), is used for determining the angle given the tangent value.
The inverse tangent function, \( \tan^{-1}(x) \), is particularly important when dealing with certain integrals. As we found from the exercise, the derivative of the inverse tangent function is \( \frac{1}{1+x^2} \). This means, to solve the definite integral \( \int_{0}^{1} \frac{1}{1+x^2} \, dx \), we essentially reverse the differentiation process by identifying that this function is the derivative of \( \tan^{-1}(x) \). Understanding these inverse relationships helps in evaluating integrals that might otherwise seem complex.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a cornerstone in calculus connecting differentiation and integration. It has two main parts, which serve as a link between the process of deriving functions and evaluating definite integrals.
In the context of the exercise, the Fundamental Theorem of Calculus allows us to evaluate \( \int_{0}^{1} \frac{1}{1+x^2} \, dx \) by using \( F(x) = \tan^{-1}(x) \) and calculating \( F(1) \) and \( F(0) \). So, knowing the antiderivative gives us a straightforward method to compute the definite integral value as \( \frac{\pi}{4} \). This illustrates beautifully how integration and differentiation are inverse operations.
- The first part tells us that an antiderivative, \( F(x) \), of a function \( f(x) \), can be found through integration.
- The second part states that if you have an antiderivative and you need to find the value of a definite integral, calculate \( F(b) - F(a) \), where \( a \) and \( b \) are the bounds of the integral.
In the context of the exercise, the Fundamental Theorem of Calculus allows us to evaluate \( \int_{0}^{1} \frac{1}{1+x^2} \, dx \) by using \( F(x) = \tan^{-1}(x) \) and calculating \( F(1) \) and \( F(0) \). So, knowing the antiderivative gives us a straightforward method to compute the definite integral value as \( \frac{\pi}{4} \). This illustrates beautifully how integration and differentiation are inverse operations.
Antiderivatives
Antiderivatives are the reverse of derivatives and play a crucial role in integration. The antiderivative of a function is another function whose derivative is the original function given.
In the solved exercise, identifying the antiderivative was key. We recognized that \( \tan^{-1}(x) \) is the antiderivative of \( \frac{1}{1+x^2} \) because its derivative gives us that exact form. Knowing how to find an antiderivative enabled us to apply the Fundamental Theorem of Calculus effectively, thus simplifying the calculation considerably. The ability to find antiderivatives unlocks many integral solutions and enhances problem-solving in calculus.
- An intuitive way to think about antiderivatives is like performing a reverse operation of differentiation.
- This concept is essential for solving indefinite integrals, which are integrals without specified limits.
In the solved exercise, identifying the antiderivative was key. We recognized that \( \tan^{-1}(x) \) is the antiderivative of \( \frac{1}{1+x^2} \) because its derivative gives us that exact form. Knowing how to find an antiderivative enabled us to apply the Fundamental Theorem of Calculus effectively, thus simplifying the calculation considerably. The ability to find antiderivatives unlocks many integral solutions and enhances problem-solving in calculus.
Other exercises in this chapter
Problem 107
Evaluate the definite integrals. $$ \int_{0}^{\pi / 4} \sin \left(x-\frac{\pi}{4}\right) d x $$
View solution Problem 108
Evaluate the definite integrals. $$ \int_{0}^{1} \sin (\pi(x+1)) d x $$
View solution Problem 110
Evaluate the definite integrals. $$ \int_{-1}^{1} \frac{4}{1+x^{2}} d x $$
View solution Problem 111
Evaluate the definite integrals. $$ \int_{0}^{1 / 2} \frac{1}{\sqrt{1-x^{2}}} d x $$
View solution